Remove given string from both start and end of a word - regex

Question

Data :
col 1
AL GHAITHA
AL ASEEL
EMARAT AL
LOREAL
ISLAND CORAL
My code :
def remove_words(df, col, letters):
regular_expression = '^' + '|'.join(letters)
df[col] = df[col].apply(lambda x: re.sub(regular_expression, "", x))
Desired output :
col 1
GHAITHA
ASEEL
EMARAT
LOREAL
ISLAND CORAL
SUNRISE
Function call :
letters = ['AL','SUPERMARKET']
remove_words(df=df col='col 1',letters=remove_letters)
Basically, i wanted remove the letters provided either at the start or end. ( note : it should be seperate string)
Fog eg : "EMARAT AL" should become "EMARAT"
Note "LOREAL" should not become "LORE"
Code to build the df :
raw_data = {'col1': ['AL GHAITHA', 'AL ASEEL', 'EMARAT AL', 'LOREAL UAE',
'ISLAND CORAL','SUNRISE SUPERMARKET']
}
df = pd.DataFrame(raw_data)

Answer 1

You may use
pattern = r'^{0}\b|\b{0}$'.format("|".join(map(re.escape, letters)))
df['col 1'] = df['col 1'].str.replace(pattern, r'\1').str.strip()
The (?s)^{0}\b|(.*)\b{0}$'.format("|".join(map(re.escape, letters)) pattern will create a pattern like (?s)^word\b|(.*)\bword$ and it will match word as a whole word at the start and end of the string.
When checking the word at the end of the string, the whole text before it will be captured into Group 1, hence the replacement pattern contains the \1 placeholder to restore that text in the resulting string.
If your letters list contains items only composed with word chars you may omit map with re.escape, replace map(re.escape, letters) with letters.
The .str.strip() will remove any resulting leading/trailing whitespaces.
See the regex demo.